Question

An aeroplane is flying horizontally along a straight line at a height of 3000 m from the ground at a speed of 160 m/s. Find the time it would take for the angle of elevation of the plane as seen from a particular point on the ground to change from 60° to 45°. Give your answer correct to the nearest second. with diagram​

Answer 1

Answer:

Approximately 8 seconds will be required for to plane to have the elevation angle as 45° from 60°.

Step-by-step explanation:

Figure attached below.

Let the height of the plane be represented by AB = 3000 m

Let the initial angle of elevation = ∠CBE = 60°

and the new angle of elevation = ∠DBE = 45°

As the height is perpendicular,

AB will be perpendicular on BE .

=> ∠ABC + ∠CBE = 90°

=> ∠ABC + 60° = 90°

=> ∠ABC = 90° - 60°

=> ∠ABC = 30°

Similarly, ∠ABD + ∠DBE = 90°

=> ∠ABD + 45° = 90°

=> ∠ABD = 90° - 45°

=> ∠ABD = 45°

Now, in ΔABC

Let the plane cover 'x' m is going from A to C

Therefore, tan ∠ABC = $\frac{x}{3000}$

=> tan 30° = $\frac{x}{3000}$

=> $\frac{1}{\sqrt3}$ =  $\frac{x}{3000}$

=> x = $\frac{3000}{\sqrt 3} * \frac{\sqrt 3}{\sqrt 3}$

=> x = 1000√3

Similarly, in ΔABD

Let the plane cover 'y' m is going from C to D

Therefore, tan ∠ABD = $\frac{x+y}{3000}$

=> tan 45° = $\frac{x+y}{3000}$

=> 1 =  $\frac{x+y}{3000}$

=> x + y = 3000

=> 1000√3 + y = 3000

=> y = 3000 - 1000√3

Now, we need to find the time taken by plane to cover the distance CD.

We know that Speed = $\frac{Distance}{Time}$

Therefore, on substituting the values,

=> 160 = $\frac{3000-1000\sqrt3}{Time}$

=> Time = $\frac{3000-1000\sqrt3}{160}$

              = $\frac{3000-1000*1.732}{160}$

              = $\frac{3000-1732}{160}$

              = $\frac{1268}{160}$

              = 7.92 sec

              ≈ 8 sec

Therefore, approximately 8 seconds will be required for to plane to have the elevation angle as 45° from 60°.

Answer 2

Answer:

8 seconds

Step-by-step explanation:

hope this helps